# 1719. Number Of Ways To Reconstruct A Tree

## Description

You are given an array pairs, where pairs[i] = [xi, yi], and:

• There are no duplicates.
• xi < yi

Let ways be the number of rooted trees that satisfy the following conditions:

• The tree consists of nodes whose values appeared in pairs.
• A pair [xi, yi] exists in pairs if and only if xi is an ancestor of yi or yi is an ancestor of xi.
• Note: the tree does not have to be a binary tree.

Two ways are considered to be different if there is at least one node that has different parents in both ways.

Return:

• 0 if ways == 0
• 1 if ways == 1
• 2 if ways > 1

A rooted tree is a tree that has a single root node, and all edges are oriented to be outgoing from the root.

An ancestor of a node is any node on the path from the root to that node (excluding the node itself). The root has no ancestors.

Example 1:

Input: pairs = [[1,2],[2,3]]
Output: 1
Explanation: There is exactly one valid rooted tree, which is shown in the above figure.


Example 2:

Input: pairs = [[1,2],[2,3],[1,3]]
Output: 2
Explanation: There are multiple valid rooted trees. Three of them are shown in the above figures.


Example 3:

Input: pairs = [[1,2],[2,3],[2,4],[1,5]]
Output: 0
Explanation: There are no valid rooted trees.

Constraints:

• 1 <= pairs.length <= 105
• 1 <= xi < yi <= 500
• The elements in pairs are unique.

## Solutions

• class Solution {
public int checkWays(int[][] pairs) {
boolean[][] g = new boolean[510][510];
List<Integer>[] v = new List[510];
Arrays.setAll(v, k -> new ArrayList<>());
for (int[] p : pairs) {
int x = p[0], y = p[1];
g[x][y] = true;
g[y][x] = true;
}
List<Integer> nodes = new ArrayList<>();
for (int i = 0; i < 510; ++i) {
if (!v[i].isEmpty()) {
g[i][i] = true;
}
}
nodes.sort(Comparator.comparingInt(a -> v[a].size()));
boolean equal = false;
int root = 0;
for (int i = 0; i < nodes.size(); ++i) {
int x = nodes.get(i);
int j = i + 1;
for (; j < nodes.size() && !g[x][nodes.get(j)]; ++j)
;
if (j < nodes.size()) {
int y = nodes.get(j);
if (v[x].size() == v[y].size()) {
equal = true;
}
for (int z : v[x]) {
if (!g[y][z]) {
return 0;
}
}
} else {
++root;
}
}
if (root > 1) {
return 0;
}
return equal ? 2 : 1;
}
}

• class Solution {
public:
int checkWays(vector<vector<int>>& pairs) {
vector<vector<bool>> g(510, vector<bool>(510));
vector<vector<int>> v(510);
for (auto& p : pairs) {
int x = p[0], y = p[1];
g[x][y] = g[y][x] = 1;
v[x].push_back(y);
v[y].push_back(x);
}
vector<int> nodes;
for (int i = 1; i <= 500; ++i) {
if (v[i].size()) {
nodes.push_back(i);
g[i][i] = 1;
}
}
sort(nodes.begin(), nodes.end(), [&](int x, int y) -> bool { return v[x].size() < v[y].size(); });
bool equal = 0;
int root = 0;
for (int i = 0; i < nodes.size(); ++i) {
int x = nodes[i];
int j = i + 1;
for (; j < nodes.size() && !g[x][nodes[j]]; ++j)
;
if (j < nodes.size()) {
int y = nodes[j];
if (v[x].size() == v[y].size()) equal = 1;
for (int z : v[x])
if (!g[y][z])
return 0;
} else
++root;
}
if (root > 1) return 0;
if (equal) return 2;
return 1;
}
};

• class Solution:
def checkWays(self, pairs: List[List[int]]) -> int:
g = [[False] * 510 for _ in range(510)]
v = defaultdict(list)
for x, y in pairs:
g[x][y] = g[y][x] = True
v[x].append(y)
v[y].append(x)
nodes = []
for i in range(510):
if v[i]:
nodes.append(i)
g[i][i] = True
nodes.sort(key=lambda x: len(v[x]))
equal = False
root = 0
for i, x in enumerate(nodes):
j = i + 1
while j < len(nodes) and not g[x][nodes[j]]:
j += 1
if j < len(nodes):
y = nodes[j]
if len(v[x]) == len(v[y]):
equal = True
for z in v[x]:
if not g[y][z]:
return 0
else:
root += 1
if root > 1:
return 0
return 2 if equal else 1


• func checkWays(pairs [][]int) int {
g := make([][]bool, 510)
v := make([][]int, 510)
for i := range g {
g[i] = make([]bool, 510)
}
for _, p := range pairs {
x, y := p[0], p[1]
g[x][y] = true
g[y][x] = true
v[x] = append(v[x], y)
v[y] = append(v[y], x)
}
var nodes []int
for i := 1; i <= 500; i++ {
if len(v[i]) > 0 {
nodes = append(nodes, i)
g[i][i] = true
}
}
sort.Slice(nodes, func(i, j int) bool {
return len(v[nodes[i]]) < len(v[nodes[j]])
})
equal := false
root := 0
for i, x := range nodes {
j := i + 1
for ; j < len(nodes) && !g[x][nodes[j]]; j++ {
}
if j < len(nodes) {
y := nodes[j]
if len(v[x]) == len(v[y]) {
equal = true
}
for _, z := range v[x] {
if !g[y][z] {
return 0
}
}
} else {
root++
}
}
if root > 1 {
return 0
}
if equal {
return 2
}
return 1
}